100/180 Quiz 3.4
End of week 7
Page 1 of 4
First Name:
Last Name:
Student-No:
Section:
Grade:
The remainder of this page has been left blank for your workings.
100/180 Quiz 3.4
End of week 7
Page 2 of 4
Very short answer questions
1. 2 marks Each part is worth 1 marks. Please write your answers in the boxes.
(a) Let y = cos(log(sin(x))). Compute
(Recall: log x = loge x = ln x.)
dy
.
dx
Answer: − sin(log(sin(x))) ·
Solution:
dy
d
= − sin(log(sin(x))) ·
(log(sin x))
dx
dx
cos x = − sin(log(sin(x))) ·
sin x
(b) If sin(x) + y 2 = ex−y+1 compute
dy
dx
at (x, y) = (0, 1).
Answer:
dy
dx
Solution: Differentiate the equation
dy
dy
x−y+1
cos(x) + 2y
=e
· 1−
dx
dx
So when x = 0, y = 1:
dy
dy
0
=e · 1−
1+2
dx
dx
dy
0=
dx
=0
cos x
sin x
100/180 Quiz 3.4
End of week 7
Name:
Student-No:
Page 3 of 4
Section:
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) Find f 0 (x) if f (x) = (x + 1)tan(x) .
Solution: We use logarithmic differentiation; so
log f (x) = log(x + 1) · tan x
Then differentiate to obtain
f 0 (x)
d
tan x
=
[log(x + 1) · tan x] = (sec x)2 · log(x + 1) +
f (x)
dx
x+1
In conclusion:
tan x
f (x) = f (x) · (sec x) · log(x + 1) +
or equivalently
x+1
tan x
tan(x)
2
= (x + 1)
· (sec x) · log(x + 1) +
x+1
0
2
√
(b) For what values of x does the derivative of arctan( x + 1) exist?
Solution: This is just the chain rule:
√
d
1
1
√
√
arctan( x + 1) =
·
dx
1 + ( x + 1)2 2 x + 1
So derivative exists provided x > −1. Hence the derivative exists on (−1, ∞).
100/180 Quiz 3.4
End of week 7
Page 4 of 4
Long answer question — you must show your work
x+2
. Compute g 0 (x) using the definition of the derivative. No
2x + 3
marks will be given for use of derivative rules, but you may use them to check
your answer.
3. 4 marks Let g(x) =
Solution:
g(x + h) − g(x)
h→0
h
1 x+h+2
x+2
= lim
−
h→0 h 2x + 2h + 3
2x + 3
1 (x + h + 2)(2x + 3) − (x + 2)(2x + 2h + 3)
= lim
h→0 h
(2x + 2h + 3)(2x + 3)
2
1 (2x + 3x + 2xh + 3h + 4x + 6) − (2x2 + 2xh + 3x + 4x + 4h + 6)
= lim
h→0 h
(2x + 2h + 3)(2x + 3)
1
−h
= lim
h→0 h (2x + 2h + 3)(2x + 3)
−1
= lim
h→0 (2x + 2h + 3)(2x + 3)
−1
=
(2x + 3)2
g 0 (x) = lim
100/180 Quiz 3.5
End of week 7
Page 1 of 4
First Name:
Last Name:
Student-No:
Section:
Grade:
The remainder of this page has been left blank for your workings.
100/180 Quiz 3.5
End of week 7
Page 2 of 4
Very short answer questions
1. 2 marks Each part is worth 1 marks. Please write your answers in the boxes.
p
dy
(a) Let y = cos(log(x)). Compute dx
.
(Recall: log x = loge x = ln x.)
sin(log(x))
Answer: − √
2x
cos(log(x))
Solution:
1
dy
d
= p
(cos(log(x)))
dx
2 cos(log(x)) dx
− sin(log(x))/x
= p
2 cos(log(x))
(b) If x2 − 2y = exy compute
dy
dx
at (x, y) = (−1, 0).
Answer:
Solution: Differentiate the equation
2x − 2y 0 = exy (y + xy 0 )
Plug in x = −1 and y = 0
2(−1) − 2y 0 = e0 (0 − y 0 )
−2 − 2y 0 = −y 0
−2 = y 0
dy
dx
= −2
100/180 Quiz 3.5
End of week 7
Name:
Student-No:
Page 3 of 4
Section:
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) Find f 0 (x) if f (x) = (x + 1)log(x)
(Recall: log x = loge x = ln x.)
Solution: We use logarithmic differentiation; so
log f (x) = log(x) · log(x + 1)
Then differentiate to obtain
f 0 (x)
d
=
[log(x) · log(x + 1)]
f (x)
dx
1
1
= · log(x + 1) + log(x)
x
x+1
In conclusion:
log(x + 1) log(x)
+
f (x) = f (x) ·
x
x+1
log(x + 1) log(x)
log(x)
+
= (x + 1)
·
x
x+1
0
√
(b) For what values of x does the derivative of arctan( x) exist?
√
Solution: Note that arctan( x) is only defined when x ≥ 0. Apply the chain
rule:
√
1
1
d
1
1
√ 2· √ =
arctan( x) =
· √
dx
1+x 2 x
1 + ( x) 2 x
So derivative exists provided x > 0. Hence the derivative exists on (0, ∞).
100/180 Quiz 3.5
End of week 7
Page 4 of 4
Long answer question — you must show your work
√
3. 4 marks Let g(x) = 1 − x. Compute g 0 (x) using the definition of the derivative. No
marks will be given for use of derivative rules, but you may use them to check
your answer.
Solution:
g(x + h) − g(x)
h→0
h
p
√
1 − (x + h) − 1 − x
= lim
h→0
h
p
p
√
√
1 − (x + h) − 1 − x
1 − (x + h) + 1 − x
·p
= lim
√
h→0
h
1 − (x + h) + 1 − x
1 1 − (x + h) − (1 − x)
= lim · p
√
h→0 h
1 − (x + h) + 1 − x
1
−h
= lim · p
√
h→0 h
1 − (x + h) + 1 − x
−1
= lim p
√
h→0
1 − (x + h) + 1 − x
−1
= √
2 1−x
g 0 (x) = lim
100/180 Quiz 3.6
End of week 7
Page 1 of 4
First Name:
Last Name:
Student-No:
Section:
Grade:
The remainder of this page has been left blank for your workings.
100/180 Quiz 3.6
End of week 7
Page 2 of 4
Very short answer questions
1. 2 marks Each part is worth 1 marks. Please write your answers in the boxes.
(a) Let y = sin(log(sin(x))). Compute
(Recall: log x = loge x = ln x.)
dy
.
dx
Answer: cos(log(sin(x))) ·
Solution:
dy
d
= cos(log(sin(x))) ·
(log(sin(x)))
dx
dx
cos x
= cos(log(sin(x))) ·
sin x
(b) If x + y + 1 = e(x
2 +y 2 )
compute
dy
dx
at (x, y) = (0, 0).
Answer:
Solution: Differentiate the equation with respect to x:
1+
dy
dy
2
2
= (2x + 2y ) · ex +y .
dx
dx
So, when x = 0 and y = 0:
1+
dy
dy
= (0 + 0 ·
) · e0
dx
dx
Therefore,
dy
= −1 .
dx
dy
dx
= −1
cos x
sin x
100/180 Quiz 3.6
Name:
End of week 7
Student-No:
Page 3 of 4
Section:
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
2
(a) Find f 0 (x) if f (x) = (sin(x) + 2)x .
Solution: We use logarithmic differentiation. First, take the logarithm
log f (x) = x2 · log sin(x) + 2
Then differentiate to obtain
d 2
cos(x)
f 0 (x)
=
x · log sin(x) + 2 = 2x · log sin(x) + 2 + x2 ·
f (x)
dx
sin(x) + 2
In conclusion:
cos(x)
2
f (x) = f (x) · 2x · log sin(x) + 2 + x ·
or equivalently
sin(x) + 2
cos(x)
x2
2
= (sin(x) + 2) · 2x · log sin(x) + 2 + x ·
sin(x) + 2
0
√
(b) For what values of x does the derivative of arctan( x − 1) exist?
Solution: This is just the chain rule:
√
d
1
1
√
arctan( x − 1) = √
·
dx
2 x − 1 1 + ( x − 1)2
1
= √
2x x − 1
So derivative exists provided x 6= 0 and x > 1. Hence the derivative exists on
(1, +∞).
100/180 Quiz 3.6
End of week 7
Page 4 of 4
Long answer question — you must show your work
2x
. Compute g 0 (x) using the definition of the derivative. No
x+5
marks will be given for use of derivative rules, but you may use them to check
your answer.
3. 4 marks Let g(x) =
Solution:
g(x + h) − g(x)
h→0
h
1 2(x + h)
2x
= lim
−
h→0 h x + h + 5
x+5
1 2(x + h)(x + 5) − 2x(x + h + 5)
= lim
h→0 h
(x + h + 5)(x + 5)
2
1 (2x + 2(h + 5)x + 10h) − (2x2 + 2x(h + 5))
= lim
h→0 h
(x + h + 5)(x + 5)
1
10h
= lim
h→0 h (x + h + 5)(x + 5)
10
= lim
h→0 (x + h + 5)(x + 5)
10
=
(x + 5)2
g 0 (x) = lim
100/180 Quiz 3.7
End of week 7
Page 1 of 4
First Name:
Last Name:
Student-No:
Section:
Grade:
The remainder of this page has been left blank for your workings.
100/180 Quiz 3.7
End of week 7
Page 2 of 4
Very short answer questions
1. 2 marks Each part is worth 1 marks. Please write your answers in the boxes.
(a) Let y = log(cos(log(x))). Compute
(Recall: log x = loge x = ln x.)
dy
.
dx
Answer:
− sin(log(x))
x cos(log(x))
Solution:
1
d
dy
=
·
(cos(log x))
dx
cos log x dx
− sin(log(x))/x
=
cos(log(x))
(b) If x2 + ex−y = y 2 + x compute
dy
dx
at (x, y) = (1, 1).
Answer:
dy
dx
= 2/3
Solution: Differentiate the equation
dy
dy
x−y
2x + e
1−
= 2y
+1
dx
dx
So when x = 1, y = 1:
dy
dy
=2
+1
2+e · 1−
dx
dx
dy
dy
3−
=2
+1
dx
dx
dy
= 2/3.
dx
0
100/180 Quiz 3.7
End of week 7
Name:
Student-No:
Page 3 of 4
Section:
Short answer questions — you must show your work
2. 4 marks Each part is worth 2 marks.
(a) Find f 0 (x) if f (x) = (ex + 1)sin(x) .
Solution: We use logarithmic differentiation; so
log f (x) = log(ex + 1) · sin x.
Then differentiate to obtain
f 0 (x)
d
ex sin x
=
[log(ex + 1) · sin x] = cos x · log(ex + 1) + x
.
f (x)
dx
e +1
In conclusion:
ex sin x
f (x) = f (x) · cos x · log(e + 1) + x
or equivalently
e +1
ex sin x
x
sin(x)
x
= (e + 1)
· cos x · log(e + 1) + x
.
e +1
0
x
(b) For what values of x does the derivative of arctan(log x) exist?
(Recall: log x = loge x = ln x.)
Solution: Just used the chain rule
d
1
1
arctan(log x) =
·
dx
1 + (log x)2 x
So this blows up when x = 0. But the original function only exist for x > 0, so
the derivative exists for all x > 0.
100/180 Quiz 3.7
End of week 7
Page 4 of 4
Long answer question — you must show your work
2x
. Compute g 0 (x) using the definition of the derivative. No
x+3
marks will be given for use of derivative rules, but you may use them to check
your answer.
3. 4 marks Let g(x) =
Solution:
g(x + h) − g(x)
h→0
h
1 2x + 2h
x
= lim
−
h→0 h x + h + 3
x+3
1 (2x + 2h)(x + 3) − 2x(x + h + 3)
= lim
h→0 h
(x + h + 3)(x + 3)
2
1 (2x + 6x + 2xh + 6h) − (2x2 + 2xh + 6x)
= lim
h→0 h
(x + h + 3)(x + 3)
1
6h
= lim
h→0 h (x + h + 3)(x + 3)
6
= lim
h→0 (x + h + 3)(x + 3)
6
=
(x + 3)2
g 0 (x) = lim